What is (\partial)/(\partial x)(-e^{kz}Awsin(kx-wt)) ?

The answer to (\partial)/(\partial x)(-e^{kz}Awsin(kx-wt)) is -e^{kz}Awcos(kx-wt)k

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(\partial)/(\partial x)(-e^{kz}Awsin(kx-wt))

\frac{\partial}{\partial x} (- e^{k z} A w sin (k x - wt))

- e^{k z} A w cos (k x - wt) k

Solution steps

\frac{\partial}{\partial x} (- e^{k z} A w sin (k x - wt))

Treat

A, k, z, w, t

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= - e^{k z} A w \frac{\partial}{\partial x} (sin (k x - wt))

Apply the chain rule:

cos (k x - wt) \frac{\partial}{\partial x} (k x - wt)

= cos (k x - wt) \frac{\partial}{\partial x} (k x - wt)

\frac{\partial}{\partial x} (k x - wt) = k

= - e^{k z} A w cos (k x - wt) k

\frac{\partial}{\partial x} (e^{4 x e^{y}})

x \to 1 - lim (\frac{2}{x} - \frac{1}{ln ( x )})

\int \frac{- x + 2}{x ^{2} + 2 x + 4} d x

\frac{d y}{d x} = e^{- 2 y} tan^{2} (x)

x^{^{''}} + x^{^{'}} - 2 x = 0

What is (\partial)/(\partial x)(-e^{kz}Awsin(kx-wt)) ?
The answer to (\partial)/(\partial x)(-e^{kz}Awsin(kx-wt)) is -e^{kz}Awcos(kx-wt)k